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Chapter 6 A.P. Chemistry. Thermochemistry. CHANGES IN. WHAT THEY BOTH HAVE IN COMMON. CHANGES IN. Total capacity for heat exchange. Internal Energy, E. Total amount of K.E. and P.E. Extensive. Enthalpy, H. State Functions. H = E + PV. q. w. D E.

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  1. Chapter 6 A.P. Chemistry Thermochemistry CHANGES IN... WHAT THEY BOTH HAVE IN COMMON CHANGES IN... Total capacity for heat exchange Internal Energy, E Total amount of K.E. and P.E. Extensive Enthalpy, H State Functions H = E + PV q w DE = + DH = DE + DPV PERSPECTIVE IS IMPORTANT!! heat …flow of energy due to temp. diff.’s DH = qp + _ work Flowing in …force x distance (moving things) FINDING THIS q = m c DT System Surroundings Flowing out + Done to VS. Calorimetry _ (Everything Else) 1. (Your Focus) Done by qsys = -(qrxn + qcal) w = -PDV 1 mole of a substance formed from its elements in std. states 2. Given the eq.’s AN EXPANDING GAS Heats of Formation. HESS’S LAW DH = Snprod.DHf – Snreact.DHf Write your own 3.

  2. 2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below. positive negative positive How much heat energy is released or absorbed when 0.500 mol of solid sodium is produced from solid sodium oxide? A.) 213 kJ is released B.) 104 kJ is absorbed C.) 104 kJ is released D.) 213 kJ is absorbed

  3. 2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below. positive negative positive With regards to the production of sodium oxide the DH value of -416 kJ refers to A.) the heat of combustion of sodium oxide B.) the heat of formation of sodium oxide C.) the heat of neutralization of sodium oxide D.) the heat of solution of sodium oxide

  4. 2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below. positive negative positive O2(g) + 4e- → 2O2-(g) Which of the following represents DHº for the reaction above? A.) x + y B.) x – 2y + z C.) x + z D.) x + 2y +2z

  5. DH = -68.1 kJ CsOH (0.125 L) (0.250 M) = 0.0313 mol CsOH HF (0.0500 L) (0.625 M) = 0.0313 mol HF Both L.R. 1:1 ratios in terms of what you need and what you have. qrxn= (175 g)(4.20 J/g°C)(2.90°C) = 2130 J qsys= -2130 J DH -2130 J/ 0.0313 mol = -68100 J/mol

  6. C2H6 (g) + 7/2 O2 (g)      2 CO2 (g) + 3 H2O (g) 1.) Find DH for this reaction by using the heats of formation from the back of the book. 2.) Use the DH you found to determine the amount of heat given off when 5.00 g of ethane reacts with 5.00 g of oxygen. 3.) What mass of each substance would be needed to release 650.0 kJ of heat? 1.) DH = [2 mol (-393.5kJ/mol) + 3 mol (-242 kJ/mol)] – [1 mol (-84.7 kJ/mol)] DH = -1428 kJ (associated with the reaction equation above) 2.) (5.00 g C2H6)(1 mol C2H6/30.0g) = 0.167 mol C2H6 ◄ L.R. (because the balanced reaction requires more oxygen than ethane and you don’t have it!) (5.00 g O2)(1 mol O2/32.0g) = 0.156 mol O2 0.156 mol O2 (-1428 kJ/3.5 mol) = -63.6 kJ 3.) O2 ► x (-1428kJ/3.5 mole) = -650 kJ x = 1.59 moles (50.9 g) C2H6► x (-1428kJ/1 mole) = -650 kJ x = 0.455 moles (13.65 g)

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